Problem: Let $V$ be a simple solid region oriented with outward normals that has a piecewise-smooth boundary surface $S$. $ \oiint_S \left[ \cos(x) \hat{\imath} + \sin(y) \hat{\jmath} + \tan(xy) \hat{k} \right] \cdot dS$ Use the divergence theorem to rewrite the surface integral as a triple integral. $ \iiint_V $ $ \, dV$
Solution: Assume we have a simple solid region $V$ oriented with outward normals, and it has a piecewise-smooth, closed boundary surface $S$. If $F$ is a continuously differentiable vector field in $\mathbb{R}^3$, then the divergence theorem says: $ \oiint_S F \cdot dS = \iiint_V \text{div}(F) \, dV$ The given surface, boundary, and vector field satisfy the conditions for the divergence theorem. We're converting a surface integral into a triple integral, so we know $F$ and we want to find $\text{div}(F)$. $\begin{aligned} F(x, y, z) &= \cos(x) \hat{\imath} + \sin(y) \hat{\jmath} + \tan(xy) \hat{k} \\ \\ \text{div}(F) &= \dfrac{\partial}{\partial x} \left[ \cos(x) \right] \\ \\ &+ \dfrac{\partial}{\partial y} \left[ \sin(y) \right] \\ \\ &+ \dfrac{\partial}{\partial z} \left[ \tan(xy) \right] \\ \\ &= -\sin(x) + \cos(y) \end{aligned}$ Therefore, the equivalent triple integral is: $ \iiint_V -\sin(x) + \cos(y) \, dV$